Question

Disregarding air resistance, what is the speed of a ball dropped from 12 feet just before it hits the ground? (Use 1 ft = 0.30 m, and use g = 9.8 m/s2.)

2.4 m/s
8.4 m/s
10.8 m/s
15.3 m/s

Answer 1

Answer: B (8.4 m/s)

Step-by-step explanation:

Answer 2

`8.4\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Let
`x` denote the displacement of the ball. It is given that
`x = 12\; {\rm ft}`. Apply unit conversion and ensure that the displacement
`x\!` of the ball is measured in meters:

`\begin{aligned}x &= (12\; {\rm ft})\, \frac{(0.30\; {\rm m})}{(1\; {\rm ft})} = 3.6\; {\rm m}\end{aligned}`.

Let
`u` denote the initial velocity of the ball, and let
`v` denote the velocity of the ball right before it hits the ground. Note that since the question states the ball was "dropped", assume that the ball was initially at rest with initial velocity
`u = 0\; {\rm m\cdot s^(-1)}`.

Under the assumptions, the acceleration
`a` of the ball will be constantly
`a = g = 9.8\; {\rm m\cdot s^(-2)`.

Rearrange the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x` to find the final velocity
`v` of the ball right before landing:

`\begin{aligned}v &= \sqrt{u^(2) + 2\, a\, x} \\ &= \sqrt{(0\; {\rm m\cdot s^(-1)})^(2) + 2\, (9.8\; {\rm m\cdot s^(-1)})\, (3.6\; {\rm m})} \\ &= 8.4\; {\rm m\cdot s^(-1)}\end{aligned}`.